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So for your ideal gas law, you are only acknowledging the air within the expansion tank, and your variables are:

P=variable you're solving for
V=?
n=?
R=?
T=coolant temperature

I think we are all in agreement that if you completely top off the expansion tank with coolant, you will get the 2bar relief pressure on the expansion tank cap?

R is the gas constant. You don't need to solve for R. You can also treat n as constant since it wouldn't change in a closed system (which it is until it starts venting).

If you set up an equation for both states, you get the following:
P1V1 = nRT1
P2V2 = nRT2

P1V1/T1 = nR
P2V2/T2 = nR

P1V1/T1 = P2V2/T2

Now you need to make some assumptions. For our purposes, it's reasonable to assume that starting temperature and pressure will be 20C and 1atm respectively. We know the final temperature gets to about 95C. Now if we assume that the system does indeed reach 3 bar, we can calculate how much the volume changes on relative terms.

That means for the pressure to reach 3 bar, the volume of air must decrease to 42% of its starting value. If we know the volume of air in the ET when cold, we can verify whether or not this actually happens by calculating how much coolant expands during operation. If it's enough to create 3 (or more) bar, then we know the cap does something.

So for your ideal gas law, you are only acknowledging the air within the expansion tank, and your variables are:

P=variable you're solving for
V=?
n=?
R=?
T=coolant temperature

I think we are all in agreement that if you completely top off the expansion tank with coolant, you will get the 2bar relief pressure on the expansion tank cap?

Apparently my posts are not long enough for some people to understand. lol.

Let me put this into context for those who are still using 2 sets of P and T or having difficulty understanding what the law says. No need to start making assumptions once you have the temperature reading. For example, no need to assume 1 atm at 20 degrees, as the atmospheric pressure depends mostly on the altitude, not temperature. You may have a situation where you fill up the system at -40 degrees near the sea level, in which case, atmospheric pressure will still be 1 atm.

P is unknown.
V is arbitrary, assume a realistic starting volume, let's say 0.5 L. It will cancel out regardless.
n is the amount of substance in moles.
R is the gas constant = 8.314 J/(K*mol), meaning it takes 8.314 Joules of energy to raise 1 mole of gas's temperature by 1 degrees Kelvin.
T is the temperature reading from the coolant temperature sensor (thanks for the correction on the location). Assume 96 degrees Celcius = 369.15 Kelvin (the latter is the unit used in the equation).

We also know that 1 mole of air occupies 22.4 L of air. 0.5 L air will have 0.02223 moles of air in it (n=0.5/22.4).

P = (0.5/22.4) * 8.314 * 369.15 / 0.5 = 137.014 Joules/L

Using 1 L = 1 dm^3 = 0.001 m^3, we have:

P = 137.014 J / 0.001 m^3 = 137,014 Pa = 137.014 kPa = 1.37014 bar...

This is the pressure of the air in the expansion tank without taking into account the expansion of the coolant.

Assuming the coolant will expand so as to fill half of the empty space in the ET, it follows, from P1/V1 = P2/V2 (special form of the ideal gas law that is valid after the maximum operating temperature is reached, after this point on, I assume T will be constant at 96 degrees Celcius), that this pressure will double to 2.74 bars...

If one does not allow the necessary space for the coolant to expand, it again follows, from P1/V1 = P2/V2 that arbitrarily large pressure can be reached. Of course, the pressure will be relieved whenever it reaches 3 bars as long as there is air left in the system, but if there is not air left, the the cap or ET will give up at some point due to high pressures of the liquid pushing against the cap.

Jesus, just get a pressure gauge and connect it to the bleeder port or modify an expansion tank cap. Then measure the max pressure.

All the math in the world will not answer the problem as there are a number of assumptions being made and many times either the assumptions are incorrect, inaccurate or you just fail to take something else into consideration.

Jesus, just get a pressure gauge and connect it to the bleeder port or modify an expansion tank cap. Then measure the max pressure.

All the math in the world will not answer the problem as there are a number of assumptions being made and many times either the assumptions are incorrect, inaccurate or you just fail to take something else into consideration.

This also assumes you do not make any basic mathematical errors.

No ****. So do you have a pressure gauge and some sort of adapter handy to do this? No? Do you even know what size threads the bleeder screw has (honestly that would be useful)? I can't just hook up any old pressure gauge to it. It has to be set up right and I don't really have the time or extra money to do it.

Apparently my posts are not long enough for some people to understand. lol.

Let me put this into context for those who are still using 2 sets of P and T or having difficulty understanding what the law says. No need to start making assumptions once you have the temperature reading. For example, no need to assume 1 atm at 20 degrees, as the atmospheric pressure depends mostly on the altitude, not temperature. You may have a situation where you fill up the system at -40 degrees near the sea level, in which case, atmospheric pressure will still be 1 atm.

P is unknown.
V is arbitrary, assume a realistic starting volume, let's say 0.5 L. It will cancel out regardless.
n is the amount of substance in moles.
R is the gas constant = 8.314 J/(K*mol), meaning it takes 8.314 Joules of energy to raise 1 mole of gas's temperature by 1 degrees Kelvin.
T is the temperature reading from the coolant temperature sensor (thanks for the correction on the location). Assume 96 degrees Celcius = 369.15 Kelvin (the latter is the unit used in the equation).

We also know that 1 mole of air occupies 22.4 L of air. 0.5 L air will have 0.02223 moles of air in it (n=0.5/22.4).

P = (0.5/22.4) * 8.314 * 369.15 / 0.5 = 137.014 Joules/L

Using 1 L = 1 dm^3 = 0.001 m^3, we have:

P = 137.014 J / 0.001 m^3 = 137,014 Pa = 137.014 kPa = 1.37014 bar...

This is the pressure of the air in the expansion tank without taking into account the expansion of the coolant.

Assuming the coolant will expand so as to fill half of the empty space in the ET, it follows, from P1/V1 = P2/V2 (special form of the ideal gas law that is valid after the maximum operating temperature is reached, after this point on, I assume T will be constant at 96 degrees Celcius), that this pressure will double to 2.74 bars...

If one does not allow the necessary space for the coolant to expand, it again follows, from P1/V1 = P2/V2 that arbitrarily large pressure can be reached. Of course, the pressure will be relieved whenever it reaches 3 bars as long as there is air left in the system, but if there is not air left, the the cap or ET will give up at some point due to high pressures of the liquid pushing against the cap.

I hope the math, and my conclusions, are clear now.

No one said that it needs to be 20°C outside for the pressure to be 1 atm. It's just a common temperature and easy to use for calculations. You made an assumption about the number of moles while WDE46 and I instead chose to make an assumption about temperature. Also, 1 mole is only 22.4L at STP. So your math essentially incorporated that assumption. That's why your numbers are similar to ours. They'd probably be even closer (or exact, I didn't check) if we used 25°C

Jesus, just get a pressure gauge and connect it to the bleeder port or modify an expansion tank cap. Then measure the max pressure.

All the math in the world will not answer the problem as there are a number of assumptions being made and many times either the assumptions are incorrect, inaccurate or you just fail to take something else into consideration.

This also assumes you do not make any basic mathematical errors.

Math > you. We have just proved that pressure can be more than 2 bars in the ET. And what were you saying?

No one said that it needs to be 20°C outside for the pressure to be 1 atm. It's just a common temperature and easy to use for calculations. You made an assumption about the number of moles while WDE46 and I instead chose to make an assumption about temperature. Also, 1 mole is only 22.4L at STP. So your math essentially incorporated that assumption. That's why your numbers are similar to ours. They'd probably be even closer (or exact, I didn't check) if we used 25°C

Ok, now that we are on the same page, I guess we can now start talking about the implications.

Jesus, just get a pressure gauge and connect it to the bleeder port or modify an expansion tank cap. Then measure the max pressure.

All the math in the world will not answer the problem as there are a number of assumptions being made and many times either the assumptions are incorrect, inaccurate or you just fail to take something else into consideration.

This also assumes you do not make any basic mathematical errors.

Actually math can make a very accurate model - that's what engineers do for a living. Now it is true we don't have all the data, and the differential equations that describe the system more accurately are beyond my abilities. But for our purposes, the approximations we make are acceptable.

Without going out and buying a pressure gauge and without sacrificing any parts, we have verified that the system can get pressurized to 3 bar with reasonable parameters. So that answered the question as to whether or not the cap makes a difference.

If the math is so offensive to you, then leave the thread. Don't take out whatever issues you may have against us.

Thanks Terra, I am familiar with the ideal gas law, I was just trying to figure out what SeanC was values he was using to get his answers. Here's what I get for being mr. streetsmart without calculations.

So, only factoring the volumetric expansion of coolant with temperature, and the ratio of air before and after coolant expansion to reach the expansion tank venting pressure, the minimum volume of air required to prevent venting through the cap is:

There will be some expansion of the block and hoses which will decrease the volumed required a small amount, my guess would be something in the range of .1L or less.

Interpretation of results:

Thinking about this value and looking at the expansion tank, I'd guess the minimum volume is very close to to amount of air that is in the expansion tank when it's properly filled. If you go to a lower pressure radiator cap, you are going to be venting pressure every time the car heats up, and then pulling a little bit of air back in every time the car cools down, and you will lose a small amount of coolant every time through evaporation. With BMW's setup, it seems like the pressure gets very close to the the limit of the expansion tank but doesn't surpass it to prevent very slow coolant losses over time.

Also, with keeping the pressure higher, and having the minimum amount of air volume required, when your engine starts to overheat, the system pressure will increase faster to reduce boiling, and the higher pressure gives a greater safety margin before there is some serious boiling and localized heating, which creates warping. Everything could be modified for a lower pressure, but you just decrease your engine's temperature safety margins, and you will need to increase the expansion tank volume prevent losing however small amount of coolant every warmup and cooldown cycle.

Damn these BMW engineers, I think they are actually pretty darn crafty, even when it comes to something as small as an expansion tank! It's a shame they didn't make the expansion tank just a little bit stronger.

Hey guys im doing an experiment on my car where i drive around with no expansion cap on, no oil cap on, a cheeseburger in the passenger seat and the fuel filler cap not screwed in as well. I will be driving under hard and soft circumstances. I will let you guys know how it turns out.

wtf, is nobody going to try and prove me wrong, and did I really not make any mathematical mistakes?

...or did everyone just really not care that much about this thread and was really bored and had time to write long posts?

Looks more or less right to me. The expansion of water will be slightly different since the rate of expansion varies with the absolute temperature, but those numbers are close enough for our purposes.

Not sure I agree with your conclusion though. Higher overall temperatures will only increase the risk of warping. If the boiling point was lower, the overall temperature will not exceed the boiling point until the coolant is converted to a gas.

wtf, is nobody going to try and prove me wrong, and did I really not make any mathematical mistakes?

...or did everyone just really not care that much about this thread and was really bored and had time to write long posts?

Umm. Why come after me? You made a point without proving it first, and when you did prove it, it was already proven? So thanks for repeating what has already been done I only wish you did it before so we didn't have to do it.

Looks more or less right to me. The expansion of water will be slightly different since the rate of expansion varies with the absolute temperature, but those numbers are close enough for our purposes.

Not sure I agree with your conclusion though. Higher overall temperatures will only increase the risk of warping. If the boiling point was lower, the overall temperature will not exceed the boiling point until the coolant is converted to a gas.

I disagree with what I put in bold. Metal warps from dissimilar expansion, not just heat. Heat metal evenly and it expands and contracts evenly with no stresses. A good example is people popping out dents with the spray from a computer duster bottle turned upside down.

Though the temperature of the coolant may not go passed the boiling point, the formation of vapor drastically reduces heat transfer from the metal of the engine to the coolant, so the engine will get much hotter than the coolant. Once a vapor bubble or pocket forms, the heat transfer rate goes from 0.58 watts per meter-kelvin to 0.016 , 97% slower. So pretty huge difference in cooling if there is a vapor pocket somewhere.

Quote:

Originally Posted by SeanC

Umm. Why come after me? You made a point without proving it first, and when you did prove it, it was already proven? So thanks for repeating what has already been done I only wish you did it before so we didn't have to do it.

lol, why are your panties in a bunch? I haven't insulted you yet, nor have I even responded to you.